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Orient the common axis of the wedge faces along $z$ and its azimuthal bisector along positive $x$. Reflection symmetries give $\bar{y}=\bar{z}=0$. In spherical coordinates, use $0\le r\le a$, $0\le \theta \le \pi$, and $-\beta /2\le \phi \le \beta /2$. The volume isSince $x=r\sin \theta \cos \phi$, the first moment isDividing by the volume locates the centre of mass:This is the centroid formula for a spherical wedge.

For the sketch, $f$ is positive and decreases from ${\lim }_{\beta \to 0^{+}}f=3\pi /16$ to ${\lim }_{\beta \to 2{\pi }^{-}}f=0$, with $f(\pi )=3/8$. Indeed, on putting $q=\beta /2$, the sign of $f^{\prime }$ is the sign of $q\cos q-\sin q$. The negative of this expression has derivative $q\sin q>0$ for $0<q<\pi$ and starts at zero, proving strict decrease.

A spherical wedge is the portion of a solid sphere between two meridional half-planes meeting at a dihedral angle $\beta$, with $0<\beta <2\pi$. Put the common axis along $z$ and its azimuthal bisector along $x$. For radius $a$, use spherical coordinates $0\le r\le a$, $0\le \theta \le \pi$, $-\beta /2\le \phi \le \beta /2$. Its volume is $2\beta a^3/3$.

For constant mass density, reflection symmetries put the centre of mass on the bisector. The first moment isDividing by volume gives $\bar{x}=3\pi a\sin (\beta /2)/(8\beta )$. It decreases from the limiting value $3\pi a/16$ for an infinitesimal wedge to zero for a full sphere.

The curved boundary is a spherical lune. Its outward vector area is $\pi a^2\sin (\beta /2){\mathbf{e}}_x$. Indeed, its $x$ component has the same angular integrals as the first moment, multiplied by $a^2$; the other components cancel by symmetry. Its scalar surface area is $2\beta a^2$, which is generally different from the magnitude of its vector area.